chapter 1 fundamental of combustion

CHAPTER 1 FUNDAMENTAL OF

COMBUSTION

Definition

Rapid oxidation of a fuel accompanied by the release of heat and/or light together with

the formation of combustion products

Fuel + oxygen Heat/light + combustion product

Fuel natural gas, fuel oil, coal and gasoline Oxygen O2 (from air) Combustion product CO2 and H2O

Combustion products

Carbon monoxide (CO) Aldehydes (e.g. H) mainly due to incomplete Unburned Fuel combustion Radicals Oxides of nitrogen (NOx) and Oxides of Sulphur (SOx)

Carbon + oxygen heat + carbon dioxide C + O2 heat + CO2

Combustion equations

CH4 + 2O2 -----> CO2 + 2H2O Reactants -----> Products + Heat

Combustion Requirement

Combustion triangle

Ignition (Energy)

Air (O2) Gas (Fuel)

Firing

(combustion)

Requirement for Successful Combustion

Fuel Air

Ignition

Molecularly Mixed Fuel

and Air

Minimum Ignition Energy

( Temperature)

Correct Amount of Fuel and Air

Residence time

Laminar Turbulent

Categories of Combustion Process

3 types of combustion process which depend on the oxygen being supplied;

Stochiometric Combustion Excess Air or Oxygen Combustion ( Fuel Lean

Combustion) Excess Fuel Combustion ( Fuel Rich Combustion)

Stoichiometric Combustion

• Stoichiometric or Theoretical Combustion is the ideal combustion process where fuel is burned completely

• chemically-correct proportion of fuel and air quantities

• no unburned fuel • no residual oxygen present in combustion products • combustion products – CO2, H2O, N2 (from air)

and heat byproduct is absent

Stoichiometric Combustion

CH4 + 2O2 CO2 + 2H2O + Heat

Excess Air or Oxygen Combustion

• Also known as ‘fuel lean’ or ‘positive excess air’ • Occurs when oxygen or air is supplied more

than the stoichiometric proportion • All fuel will react with O2 however excess left

over O2 will remain in the product. • Combustion products - CO2, H2O, O2 (from air),

N2 (from air) and heat

Excess Air or Oxygen Combustion

CH4 +3O2+ignition CO2 + 2H2O + O2 + Heat

Excess Fuel Combustion

• Also known as ‘fuel rich’ or ‘negative excess air’ • Occurs when fuel is supplied more than the

stoichiometric proportion. • Insufficient amount of oxygen or air available to

burn in the fuel-rich mixture caused incomplete combustion

• All fuel not completely react with O2 thus produce CO in the product

• Combustion products – CO, H2O (from air), N2 (from air) and Heat

Excess Fuel Combustion

CH4 + O2 + ignition CO + H2O + H2 + Heat

Incomplete Combustion

Incomplete combustion should be avoided due to economy and safety factor; Economy – Excess fuel which is not being used in combustion is wasted. This will reduce the combustion efficiency as it will use up the heat produce from the combustion reaction which is to be used in heating or other process. Safety – CO is a toxic gas that is dangerous to health and environment

Excess air

• Usually combustion chamber incorporates a modest amount of excess air - about 10 to 20% more than what is needed to burn the fuel completely.

• 150% theoretical air = 50% excess air • If an insufficient amount of air is supplied to the

burner, unburned fuel, soot, smoke, and carbon monoxide exhausts from the boiler - resulting in heat transfer surface fouling, pollution, lower combustion efficiency, flame instability and a potential for explosion.

Combustion Terminology

• Reactants • Fuels • Oxidizer • Products • Complete combustion • Inerts (Nitrogen, Argon) • Modeling Combustion Air • Fuel/Air Ratio • Theoretical Air • Dew point • Adiabatic Flame Temperature

Modeling Combustion Air

• Air consist N2, O2, CO2, Argon and traces of other gas

• Only O2 is reactive component in air, N2 is inert (un-reactive)

• Molar basis, air consists of 79% N2 and 21% O2.

• 1 mole of air contain 0.79 mole of N2 and 0.21 mole of O2

• Thus each one mole of O2 needed to oxidized hydrocarbon is accompanied by 79/21 = 3.76 moles of N2

Combustion equations

CH4 + 2(O2 +3.76N2) -----> CO2 + 2H2O +7.52N2

CH4 + 2O2 -----> CO2 + 2H2O

Fuel/Air Ratio

• The standard measure of the amount of air used in a combustion process is the Fuel/Air Ratio (FA)

• Defined as FA = mass of fuel/mass of air

FA = mfuel = 1 [kmol](12+4)[kg/kmol] = 0.058 kg-fuel mair 2(4.76)[kmol] 29[kg/kmol] kg-air

*Mair = 29kg/kmol

Theoretical air

• The minimum amount of air for complete combustion • Product contain no oxygen • If supply is less, CO may be present in product • Normal practice to supply more than the theoretical air • Excess air will result in O2 appear in products

CH4 + 2O2 CO2 + 2H2O

• 1 molecule CH4 requires 2 molecules O2

• Theoretical O2 is 2Nm3 per 1 Nm3 of CH4

Theoretical air requirement Because oxygen proportion in air is about 21 percent at volume basis, theoretical air requirement is calculated as follows. Combustion equation:

CH4 + 2O2 CO2 + 2H2O

So theoretical air requirement for 1 Nm3 of CH4 is 2Nm3 x 100/21 = 9.52 Nm3/Nm3

Problem 1

Air/gas Ratio

Air/gas ratio is defined as the proportion of actual air volume to theoretical air volume, as follows: Air/gas ratio = Aa / At

where Aa is actual air volume of combustion At is theoretical air requirement of fuel Excess air ratio is calculated as follows: Excess air ratio = (air/gas ratio) – 1 = (Aa /At)-1

Dew point

• Minimum temperature below which the water vapour in the combustion products will condense

• Dew temperature of natural gas ~ 60oC

Adiabatic Flame Temperature

• The maximum achievable temperature is reached when the furnace or combustion chamber is well insulated

• Typical Tad ~ 1950oC

• Tad is calculated from energy balance

Combustion Stoichiometry

CH4 + 2(O2 + 3.76 N2) CO2 + 2H2O + 7.52 N2

C2H6 + a(O2 + 3.76 N2) bCO2 + cH2O + dN2

C3H8 + a(O2 + 3.76 N2) bCO2 + cH2O + dN2

C4H10 + 6.5(O2 + 3.76 N2) 4CO2 + 5H2O + 24.4N2

Example 1

Find the theoretical air requirement for propane combustion Equation : Theoretical O2 : Theoretical air :

Example 2 Calculations of theoretical oxygen and air requirements for natural gas supplied before ’95 shown below;

Symbol

Combustion equation

Component (Vol %)

(A)

O2 Required (m3 / m3)

(B)

Total O2

CH4

CH4 + 2 O2 = CO2 + 2 H2O

84.75

2.0

1.695

C2H6

C2H6 + 3.5 O2 = 2 CO2 + 2 H2O

10.41

3.5

0.364

C3H8

C3H8 + 5 O2 = 3 CO2 + 4 H2O

0.98

5.0

0.049

C4H10

C4H10 + 6.5 O2 = 4 CO2 + 5 H2O

0.11

6.5

0.007

N2

Non combustion (no effect)

0.19

0.0

-

CO2

Non combustion (no effect)

3.36

0.0

-

TOTAL

100.00

-

2.115

Example 2

Theoretical oxygen requirement per unit of volume is 2.115 m3/m3.

Air requirement is calculated using 21 per cent of oxygen content in air as follows: 2.115 m3 x 100/21 = 10.07 m3/m3

Example 3

Symbol

Combustion equation

Component

(Vol %) (A)

O2 Required (m3 / m3)

(B)

Total O2

C3H8

C3H8 + 5 O2 = 3 CO2 + 4 H2O

30.00

5.0

1.500

C4H10

C4H10 + 6.5 O2 = 4 CO2 + 5 H2O

70.00

6.5

4.550

N2

Non combustion (non effect)

0.19

0.0

-

CO2

Non combustion (non effect)

3.36

0.0

-

TOTAL

100.00

-

6.050

Theoretical oxygen required is 6.050 m3/m3

Theoretical air required is 28.810 m3/m3

Calculations of theoretical oxygen and air requirements for LPG

Question?

Exercises

Now please complete the solution for Example 4 (page 13)

THE END

Thank You for the Attention